Let $X$ be a metrizable space and $\{x_n\}$ be a sequence in $X$. Suppose the sequence of Dirac measures $\delta_{x_n} \xrightarrow{w} P$ where $P$ is some probability measure. Prove that $P = \delta_x$ for some $x \in X$.

I am not sure how to go about proving this exactly...

The definition of weak convergence I am using is the standard one: so $\delta_{x_n} \xrightarrow{w} P$ happens when for every continuous and bounded $f: X \xrightarrow{} \mathbb{R}$ we have $\int f d\delta_{x_n} \xrightarrow{} \int f dP$.


We can follow (and detail) the following steps.

  1. Fix an integer $r$ and construct a continuous and bounded function $f$ such that $f(x)=1$ if $x$ belongs to the closure of $F_r:=\bigcap_{n\geqslant r}\{x_n\}$.
  2. The definition of weak convergence implies that for each $r$, $\mathbb P(\overline{F_r})=1$.
  3. In particular, $\bigcap_r\overline{F_r}$ is non empty. It remains to prove that this set cannot contain two distinct elements

I saw this exercise in Billingsley's book (Convergence of Probability Measures). Here is my solution.

First, if $x_n$ has a convergent subsequence, which we identify by $x_{n_k}\to x$, then it is easy to see that $x_{n_k}\to x \Rightarrow [\delta_{x_{n_k}}\Rightarrow \delta_x]$. Since, $\delta_{x_{n_k}}\Rightarrow P$, we must have $P=\delta_x$.

Now, suppose that $x_n$ does not have any convergent subsequence. Then, it does not have a limit. In particular, none of $x_k$'s can be its limit, hence, for every given $x_k$, there is an $\epsilon_k>0$, such that for a subsequence $\{x_{n_i}\}_{i=1}^\infty$ the metric ball $B(x_k,\epsilon_k)$ does not contain any of $x_{n_i}$'s (since $d(x_k,x_{n_i})\geq\epsilon_k,\forall i$).

Since the ball $B(x_k,\epsilon_k)$ is open, we have, by Portmanteau's theorem that, $$ 0=\liminf_{n\to\infty}\delta_{x_n}(B(x_k,\epsilon_k))\geq P(B(x_k,\epsilon_k)). $$ Next, let $\displaystyle F=\bigcup_n \{x_n\}$. Observe that $F$ does not have a limit point (otherwise, we would be able to identify a subsequence converging to this limit point, contradicting with the hypothesis that $x_n$ does not have any convergent subsequence), hence $F$ must be closed. Therefore, one more application of Portmanteau's theorem gives us, $$ 1=\limsup_{n\to\infty}\delta_{x_n}(F)\leq P(F)\implies P(F)=1. $$ Finally, since $\displaystyle F\subset \bigcup_n B(x_n,\epsilon_n)$, and since, $$ 1=P(F)\leq P\left(\bigcup_n B(x_n,\epsilon_n)\right)\leq \sum_{n}P(B(x_n,\epsilon_n))=0, $$ we get a contradiction.


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